What is the working load of a compression member with effective sectional area as 7500 mm² and design compressive stress of 300 N/mm² ? (Take factor of safety (FoS) = 1·5)
- A2250 kN
- B1500 kNCorrect
- C750 kN
- D562·5 kN
Solution
Concept
Working load = design strength ÷ factor of safety
Formula
Pworking = Pd / FoS
Step-by-step solution
1. Design strength = Ae × fcd = 7500 × 300 = 2.25 × 10⁶ N = 2250 kN. 2. Working load = 2250 / 1.5 = 1500 kN.
Explanation
The design compressive strength already includes the partial safety factor for material; the working load is obtained by dividing it by the further factor of safety.
Why the other options are wrong
2250 kN is the design strength, not the working load. 750 and 562.5 kN use a factor of 3 and 4.
Shortcut
Working load = A fcd / FoS.
Common mistake
Giving the design strength.
Correct answer: B) 1500 kN